Question:
If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Updated On: Jan 13, 2026
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Solution and Explanation
PR = QR
\(\sqrt{(5-0)^2+(-3-1)^2}=\sqrt{(0-x)^2+(1-6)^2}\)
\(\sqrt{(5)^2+(-4)^2}=\sqrt{(-x)^2+(-5)^2}\)
\(\sqrt{25+16}=\sqrt{x^2+25}\)
41=\(x^2+25\)
\(x^2=16\)
\(x=\pm 4\)
Therefore, point R is (4,6) or (-4,6)
When point R is (4,6)
PR= \(\sqrt{(5-4)^2+(-3-6)^2}=\sqrt{(1)^2+(-9)^2}=\sqrt{1+81}=\sqrt{82}\)
QR= \(\sqrt{(0-4))^2+(1-6)^2}=\sqrt{(4)^2+(-5)^2}=\sqrt{16+25}=\sqrt{41}\)
When point R is (-4,6)
PR= \(\sqrt{(5-(-4))^2+(-3-6)^2}=\sqrt{(9)^2+(-9)^2}=\sqrt{81+81}=\sqrt{162}\)
QR= \(\sqrt{(0-(-4))^2+(1-6)^2}=\sqrt{(4)^2+(-5)^2}=\sqrt{16+25}=\sqrt{41}\)
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